1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Prove using the Triangle Inequality

  1. Sep 23, 2011 #1
    1. The problem statement, all variables and given/known data

    Show that:

    (|x+y|)/(1+|x+y|) ≤ ((|x|)/(1+|x|)) + ((|y|)/(1+|y|))

    2. Relevant equations

    You are given the triangle inequality:

    |x+y| ≤ |x| + |y|

    3. The attempt at a solution

    (This is done from the result, as I haven't been able to find the starting point)

    (|x+y|)/(1+|x+y|) ≤ (|x|(1+|y|)+|y|(1+|x|))/((1+|x|)(1+|y|))

    (|x+y|)/(1+|x+y|) ≤ (|x|+2|x||y|+|y|)/(1+|x|+|y|+|x||y|)

    This doesn't seem to go anywhere. I also tried flipping the whole thing to get:


    but this doesn't seem to lead anywhere either...

    I'm not sure how to go about this problem.
  2. jcsd
  3. Sep 23, 2011 #2
    As a first step, try starting by applying the triangle inequality to the numerator of the left-hand side. Next, what can you say about [itex] 1 + |x+y|[/itex] and [itex] 1+|x|[/itex]? What does this imply about [itex] \frac{1}{1+|x+y|}[/itex] and [itex] \frac{1}{1+|x|}[/itex]?
  4. Sep 24, 2011 #3
    so, since 1+|x+y| [itex]\geq[/itex] 1+|x|,

    [itex]\frac{1}{1+|x+y|}[/itex] [itex]\leq[/itex] [itex]\frac{1}{1+|x|}[/itex]


    [itex]\frac{|x+y|}{1+|x+y|}[/itex] [itex]\leq[/itex] [itex]\frac{|x|}{1+|x+y|}[/itex] + [itex]\frac{|y|}{1+|x+y|}[/itex]


    [itex]\frac{|x|}{1+|x+y|}[/itex] + [itex]\frac{|y|}{1+|x+y|}[/itex] [itex]\leq[/itex] [itex]\frac{|x|}{1+|x|}[/itex] + [itex]\frac{|y|}{1+|x+y|}[/itex]

    Can a similar argument now simply be extended for 1+|y|?
  5. Sep 24, 2011 #4
    You're on the right track. Prove the inequality is true for the denominator, then you can easily prove it is true for the numerator.
  6. Sep 24, 2011 #5
    Actually, this isn't true: take x=1 and y=-1 for a counterexample.

    I just noticed a trick that makes this much easier:

    \frac{a}{1+a} = \frac{1+a-1}{1+a} = \frac{1+a}{1+a} - \frac{1}{1+a} = 1 - \frac{1}{1+a}\; .

    Try using this trick on the LHS. Then you can use the triangle inequality (which will give you [itex] 1+|x+y| \leq 1 + |x| +|y|[/itex]) to compare [itex] -\frac{1}{1+|x+y|}[/itex] and [itex]-\frac{1}{1+|x|+|y|}[/itex]. Then you can use the first trick backwards, which should lead to the result. And hopefully I haven't made any mistakes with my inequalities!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook